确定矩阵[][]的对称性

Question

我试图找出一个矩阵是否在不同的方向上是对称的(horitzonzal/垂直的或两者都是)，并找到了本教程https://www.geeksforgeeks.org/check-horizontal-vertical-symmetry-binary-matrix/，并尝试了我的矩阵，但它似乎不能正确地工作。下面是我的当前代码和我得到的输出：

示例矩阵：

111
144
144

我通过以下方式调用本教程中的方法：

 // test for symmetrie 
    checkHV(matrix, rows, colums);

在这种情况下，行和列都是3。在我看来，它应该输出NO，但是当前的输出是垂直的(所有矩阵都是.)。为什么会这样呢？我如何修改代码，使其对我正确工作呢？谢谢!

下面是教程中的代码：

 static void checkHV(int [][]arr, int N,
                int M)
{

// Initializing as both horizontal 
// and vertical symmetric.
boolean horizontal = true;
boolean vertical = true;

// Checking for Horizontal Symmetry. 
// We compare first row with last
// row, second row with second
// last row and so on.
for (int i = 0, k = N - 1; 
         i < N / 2; i++, k--)
{

    // Checking each cell of a column.
    for (int j = 0; j < M; j++)
    {

        // check if every cell is identical
        if (arr[i][j] != arr[k][j])
        {
            horizontal = false;
            break;
        }
    }
}

// Checking for Vertical Symmetry. We compare
// first column with last column, second xolumn
// with second last column and so on.
for (int i = 0, k = M - 1;
         i < M / 2; i++, k--)
{

    // Checking each cell of a row.
    for (int j = 0; j < N; j++)
    {
        // check if every cell is identical
        if (arr[i][j] != arr[k][j])
        {
            horizontal = false;
            break;
        }
    }
}

if (!horizontal && !vertical)
    System.out.println("NO");

else if (horizontal && !vertical)
    System.out.println("HORIZONTAL");

else if (vertical && !horizontal)
    System.out.println("VERTICAL");

else
    System.out.println("BOTH");
}

编辑：在更改horitz周= true = true for垂直=true时，代码仍然不能很好地处理矩形矩阵，所以f.e。4*2，给了我一个超出界限的数组。矩阵：

1112
2212

行=2，列=4。

Answer 1

试试这个：

 static void checkHV(int [][]arr, int N,
            int M)
{

// Initializing as both horizontal 
// and vertical symmetric.
boolean horizontal = true;
boolean vertical = true;

// Checking for Horizontal Symmetry. 
// We compare first row with last
// row, second row with second
// last row and so on.
for (int i = 0, k = N - 1; 
         i < N / 2; i++, k--)
{

    // Checking each cell of a column.
    for (int j = 0; j < M; j++)
    {

        // check if every cell is identical
        if (arr[i][j] != arr[k][j])
        {
            horizontal = false;
            break;
        }
    }
}

// Checking for Vertical Symmetry. We compare
// first column with last column, second xolumn
// with second last column and so on.
for (int i = 0, k = M - 1;
         i < M / 2; i++, k--)
{

    // Checking each cell of a row.
    for (int j = 0; j < N; j++)
    {
        // check if every cell is identical
        if (arr[j][i] != arr[j][k])
        {
            vertical = false;
            break;
        }
    }
}

if (!horizontal && !vertical)
    System.out.println("NO");

else if (horizontal && !vertical)
    System.out.println("HORIZONTAL");

else if (vertical && !horizontal)
    System.out.println("VERTICAL");

else
    System.out.println("BOTH");
}

Answer 2

public static boolean isSymmetricHorizontally(int[][] matrix) {
    for(int rowL = 0, rowH = matrix.length - 1; rowL < rowH; rowL++, rowH--)
        for(int col = 0; col < matrix[0].length; col++)
            if(matrix[rowL][col] != matrix[rowH][col])
                return false;
    
    return true;
}

public static boolean isSymmetricVertically(int[][] matrix) {
    for(int colL = 0, colH = matrix[0].length - 1; colL < colH; colL++, colH--)
        for(int row = 0; row < matrix[0].length; row++)
            if(matrix[row][colL] != matrix[row][colL])
                return false;
    
    return true;
}